If a quotient group is isomorphic to another group, can I "solve" for the latter Let $N\triangleleft{A}$ and ${A}/{N}\cong{B}$ for some non-trivial groups $A$ and $B$. Then, is the expression $N\cong{A}/{B}$ valid in general? If not, are there any conditions for its validity?
In particular, I have in hand the expressions:
$${{\pi }_{1}}{{\left( H \right)}_{G}}\triangleleft {{\pi }_{1}}\left( H \right)\quad\text{and}\quad{{{\pi }_{1}}\left( H \right)}/{{{\pi }_{1}}{{\left( H \right)}_{G}}}\;\overset{\mathbf{Grp}}{\mathop{\cong }}\,{{\pi }_{1}}\left( G \right)$$
where $H\le G$ is a topologically closed subgroup of the Lie group $G$ and ${{\pi }_{1}}{{\left( H \right)}_{G}}$ is the set of homotopy groups of loops that are trivial when embedded in $G$, i.e. the kernel of $i_*$, where $i\colon{H}\to{G}$ is the inclusion map. I would like from these to infer that:
$${{\pi }_{1}}{{\left( H \right)}_{G}}\overset{\mathbf{Grp}}{\mathop{\cong }}\,{{{\pi }_{1}}\left( H \right)}/{{{\pi }_{1}}\left( G \right)}$$
whithout any assumptions on the connectedness of the relevant groups, except from:
$${{\pi }_{0}}\left( H \right)\overset{\mathbf{Set}}{\mathop{\cong }}\,{{\pi }_{0}}\left( G \right)$$
If these conditions are not enough to get to the desired result, what other assumptions shall I include?

So, for general groups the first part of the answer is negative. What about the part involving the homotopy groups? Is there any loophole there? Even if I have to relax the conditions on connectedness of $G$ and $H$ and assume that one of them (therefore the other, also) is path-connected? This would imply, if I am correct, that the fundamental groups Abelianize.
 A: To answer your general question from your first paragraph: no, it's not valid. And the reason is that $B$ may not be a subgroup of $A$ at all, so the expression "$A/B$" wouldn't even make sense. Here's a quick example: $A=\mathbb{Z}$ and $N=2\mathbb{Z}$. Then $A/N=\mathbb{Z}/2\mathbb{Z}=\mathbb{Z}_2$, and there's no such subgroup in $A$.
A: I am not familiar with homotopy theory, so please correct me if this is an irrelevant counterexample.  From a purely group-theoretic standpoint, we do not have $A/N \cong B \implies N \cong A/B$.
For instance, there is a normal subgroup isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$ in $S_4$, namely $\{ \text{id}, \ (12)(34), \ (13)(24), \ (14)(23) \}$.  One can check that $S_4 / \mathbb{Z}_2 \times \mathbb{Z}_2 \cong S_3$ (there are two non-isomorphic groups of order $6$, and this quotient group is not abelian); see this thread for reference.  However, we do not have $S_4/S_3 \cong \mathbb{Z}_2 \times \mathbb{Z}_2$.  Indeed, $S_3$ isn't even normal in $S_4$.
